I am stuck with showing that this function is differentiable at $x=0$ but not at $x=1$.

$f(x) = \begin{cases} x^2, & \text{if x}\in{\mathbb{Q}}\\ x^3, & \text{if x}\notin{\mathbb{Q}}\\ \end{cases}$

So for $x=0$:

Case 1: $h \in \mathbb{Q}$

$\lim\limits_{h \to 0}{\frac{f(0+h)-f(0)}{h}} = \lim\limits_{h \to 0}{\frac{f(h)}{h}} = \lim\limits_{h \to 0}{\frac{h^2}{h} = \lim\limits_{h \to 0}{ h} = 0} $

Case 2: $h \notin \mathbb{Q}$

$\lim\limits_{h \to 0}{\frac{f(0+h)-f(0)}{h}} = \lim\limits_{h \to 0}{\frac{f(h)}{h}} = \lim\limits_{h \to 0}{\frac{h^3}{h} = \lim\limits_{h \to 0}{ h^2} = 0} $

I am unsure on how to continue from here. I thought I need to show that for $x=0$ the right and left hand side limits are the same, and for $x=1$ they are different.


For $\;x=1\;$ we'd get

$$\frac{f(1+h)-f(1)}h=\begin{cases}&\frac{(1+h)^2-1}h=\frac{2h+h^2}h=(2+h)\xrightarrow[h\to0]{}2,&h\in\Bbb Q\\{}\\ &\frac{(1+h)^3-1}h=\frac{3h+3h^3+h^3}h=(3+3h+h^2)\xrightarrow[h\to0]{}3,&h\notin\Bbb Q\end{cases}$$

  • 2
    $\begingroup$ Oh I see... I simply forgot to do the same thing for $x=1$. Here we can see that they are not the same. Thanks! $\endgroup$ – ffritz Apr 20 '17 at 12:53
  • $\begingroup$ @Warhost Indeed...yet in zero the limits are equal and thus the function's differentiable there. No need to do one-sided limits in this case. $\endgroup$ – DonAntonio Apr 20 '17 at 12:54


for $x=1$ the limit of the quotient goes to $2$ for rational $h$ and goes to $3$ for irrational $h$.


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